Notes about training GAN

• Try more complex generator
• ...so simpler the discriminator
• A "stronger" dataset: the style of the distribution should be more highlighted
  

G' and G''

Exercises 7.41 and 8.14 from Rubinstein's book Polymer Physics. From Boltzmann superposition principle, we have $$\sigma (t)={\int }_{-\mathrm{\infty }}^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })\mathrm{d}{t}^{\prime }$$ $$\gamma (t)={\gamma }_0\exp (I\omega t)$$ $$\begin{array}{rl}\sigma (t)& =I\omega {\gamma }_0{\int }_{-\mathrm{\infty }}^{t}G(t-t^{\prime })\exp (I\omega t^{\prime })\mathrm{d}{t}^{\prime }\\ & =I\omega {\gamma }_0\exp (I\omega t){\int }_0^{\mathrm{\infty }}G(\tau )\exp (-I\omega \tau )\mathrm{d}\tau \\ & :=\gamma (t)I\omega \hat{G}(\omega )\end{array}$$ We therefore have $$I\omega \hat{G}=I\omega {\mathcal{F}}^{\cos }\{G\}+\omega {\mathcal{F}}^{\sin }\{G\}\equiv I{G}^{\prime \prime }+G^{\prime }$$ ${\mathcal{F}}^{\cos }$ and ${\mathcal{F}}^{\sin }$ represent Fourier cosine and sine transform. Take ${\mathcal{F}}^{\cos }(1)=\delta (\omega )$ and ${\mathcal{F}}^{\sin }(1)=1/\omega$ (note the normalization factor of $\mathcal{F}$), let $G(t)=G_{eq}+G_1(t)$: $$\begin{array}{rl}{G}^{\prime }& =\omega {\mathcal{F}}^{\sin }(G_1+G_{eq})=G_{eq}+\omega {\mathcal{F}}^{\sin }(G_1(t))\\ G^{\prime \prime }& =\omega {\mathcal{F}}^{\cos }(G_1+G_{eq})=\omega {\mathcal{F}}^{\cos }(G_1(t))+\omega \delta (\omega )=\omega {\mathcal{F}}^{\cos }(G_1(t))\end{array}$$ $G^{\prime }$ and $G^{″}$ are defined as the Fourier sine and cosine transform of stress relaxation function $G(t)$: $$\begin{array}{rl}{G}^{\prime }& :=\omega {\int }_0^{\mathrm{\infty }}G(t)\sin \omega t\mathrm{d}t\\ G^{\prime \prime }& :=\omega {\int }_0^{\mathrm{\infty }}G(t)\cos \omega t\mathrm{d}t\end{array}$$ Let $G(t)$ be $$G(t)\sim {\left(\frac{t}{{\tau }_0}\right)}^{-\frac{1}{2}}\exp (-\frac{t}{{\tau }_R})$$ G' i.e., the cosine transform is thus $$G^{\prime \prime }/\omega ={\int }_0^{\mathrm{\infty }}\mathrm{d}t{\left(\frac{t}{{\tau }_0}\right)}^{-\frac{1}{2}}\exp (-\frac{t}{{\tau }_R})\cos \omega t$$ Since ${\tau }_R=N^2{\tau }_0$, we have (let ${t^{\prime }}^2{\tau }_R=t$) $$\begin{array}{rl}{G}^{\prime \prime }/\omega & =2\frac{{\tau }_R}{N}{\int }_0^{\mathrm{\infty }}\mathrm{d}{t}^{\prime }\exp (-{t^{\prime }}^2)\cos \omega {\tau }_R{t^{\prime }}^2\\ & =2\frac{{\tau }_R}{N}\mathrm{Re}{\int }_0^{\mathrm{\infty }}\mathrm{d}{t}^{\prime }\exp (-{t^{\prime }}^2)\exp \left(I\omega {\tau }_R{t^{\prime }}^2\right)\\ & =2\frac{{\tau }_R}{N}\mathrm{Re}{\int }_0^{\mathrm{\infty }}\mathrm{d}{t}^{\prime }\exp (-(1-I\omega {\tau }_R){t^{\prime }}^2)\end{array}$$ Since $t^2=(-t{)}^2$, extending the range of last integral to $(-\mathrm{\infty },\mathrm{\infty })$ yields a Gaussian integral with complex argument, the result is straightforward:
$$\begin{array}{rl}{G}^{\prime \prime }/\omega & =\frac{{\tau }_R}{N}\mathrm{Re}\sqrt{\frac{\pi }{1-I\omega {\tau }_R}}\\ & =\frac{{\tau }_R}{N}\mathrm{Re}\sqrt{\pi }\frac{\sqrt{1+I\omega {\tau }_R}}{1+(\omega {\tau }_R{)}^2}\\ & =\frac{{\tau }_R}{N}\mathrm{Re}\sqrt{\pi }\frac{1}{1+(\omega {\tau }_R{)}^2}\sqrt{\sqrt{1+(\omega {\tau }_R{)}^2}}\times & \\ & (\cos (\frac{1}{2}\arctan \omega {\tau }_R)+I\sin (\frac{1}{2}\arctan \omega {\tau }_R))\\ & =\frac{{\tau }_R}{N}\sqrt{\pi }\frac{1}{1+(\omega {\tau }_R{)}^2}\sqrt{\sqrt{1+(\omega {\tau }_R{)}^2}}\cos (\frac{1}{2}\arctan \omega {\tau }_R)\end{array}$$ With $\cos (\arctan (x))=\frac{1}{1+x^2}$ and $2\cos (x/2{)}^2-1=\cos (x)$, we finally have $$G^{\prime \prime }=\omega {\tau }_R\sqrt{\frac{\pi }{2}}\sqrt{\frac{\sqrt{1+(\omega {\tau }_R{)}^2}+1}{1+(\omega {\tau }_R{)}^2}}$$ The factor $\frac{\pi }{2}$ comes from the definition of Fourier cosine transform.
P.S. It is more convinient (non-restrictively) to treat the integral $$t^{-\alpha }\exp (-t)\exp (Iwt)$$ with $0<\alpha <1$ as Gamma function with (complex) coefficient in $\exp$: $${\int }_0^{\mathrm{\infty }}\mathrm{d}t{t}^{-\alpha }\exp (-(1-I\omega )t)=(1-I\omega t{)}^{\alpha -1}\mathrm{\Gamma }(1-\alpha )$$ We may use Hypergeometric function ${}_p{F}_q$ for more genernal cases, i.e., $t^{-\alpha }\exp (-t^{\beta })\exp (Iwt)$